Spectrum of a product

Question

Let $A$ be a unital $C^{*}$-algebra. I am trying to show that if $a,b\in A$ are positive elements, then the spectrum of $ab$ is contained in the positive real numbers. I know that in the commutative case, $\sigma(ab)\subseteq\sigma(a)\sigma(b)$ for any elements $a,b$ but here there is no assumption of commutativity.

Answer 1

Use $$ \sigma(ab) \cup \{0\} = \sigma(ba) \cup \{0\} .$$

See Do $X(X'X)^{-1}(X'X)^{-1}X'$ and $(X'X)^{-1}$ have the same non-zero eigenvalues? for the proof of this (although if $a$ or $b$ is invertible, as it would be if $a$ and $b$ are positive, then $\sigma(ab) = \sigma(ba)$ can be proved much more easily.)

Then $\sigma(ab) = \sigma(a^{1/2} a^{1/2} b) = \sigma(a^{1/2} b a^{1/2})$, and $a^{1/2} b a^{1/2}$ is positive.